Find all solutions of the equation in the interval [0,2π)

Question

√3 tanθ +1 = 0

Write your answer in radians in terms of π.

If there is more than one solution, separate them with commas.

Answer 1

√3•tanx + 1 = 0

tanx = -1/√3

Then we can draw a right triangle, and take positive tan value firstly. Tan is opp/adj.

After drawing that triangle we can easily identify that angle, because of the relationship between the sides of that triangle (one of the special right triangles). The angle is 30° or π/6

We know tan is negative in 2nd and 4th quadrants. From the periods of the tan, angle could be (π-π/6) which is 5π/6 or (2π-π/6) which is 11π/6

Answer 2

From √3 tan(θ) + 1 = 0, we arrive at √3 tan(θ) = -1. Then, tan(θ) = -1/√3. Since tan(θ) = sin(θ)/cos(θ), we have that sin(θ)/cos(θ) = -1/√3. Then, -√3 sin(θ) = cos(θ). Squaring both sides, 3 sin²(θ) = cos²(θ). Here we use the identity 1 - sin²(θ) = cos²(θ) such that, by substitution, 3 sin²(θ) = 1 - sin²(θ). This implies that 4 sin²(θ) = 1, which in turn implies sin²(θ) = 1/4, or sin(θ) = ±1/2. This is easily recognized, in quadrant 1, as θ = π/6. On the interval [0, 2π), we have a few more solutions. All of the solutions are {θ | θ ∈ {π/6, 5π/6, 7π/6, 11π/6}}.

We must be careful here and consider the original problem. Our solution set is a solution set for sin(θ) = ±1/2. From the original problem, we can easily see that tan(θ) < 0 must be true. We know this to be true for quadrants 2 and 4. Thus, we must restrict our solution set to only allow the solutions that are in quadrants 2 or 4, namely {θ | θ ∈ {5π/6, 11π/6}}.